PHYSICS SET UP

Ray Williams

April 10, 2013

TRIDENT UNIVERSITY Module 1 - Case Displacement, Velocity, and Acceleration Ray Williams 4/23/2013

Module 1 - Case

Displacement, Velocity, and Acceleration

1. A car is driving along a straight section of highway between Oklahoma City and Tulsa. At one point in time, the car is 50 miles from Tulsa. At a later point in time, it's 40 miles from Tulsa. What is its displacement?

SHOW WORK :

Displacement is the change in vector position, that is to say, it's simply the distance traveled, with a direction attached to it (a vector).

At point A, it was 50 miles from Tulsa, and at point B, it was 40 miles from Tulsa. This means it travelled 50-40=10 miles in total.

If we take point B, Tulsa, as our reference point, we can say that the car's displacement was 10 miles [towards Tulsa]. Alternatively, if point A, Oklahoma, was our reference point, it would be 10 miles [away from Oklahoma].

ANSWER:10 miles

2. An ant is placed on a piece of graph paper. It crawls 3 cm on an azimuth (course angle) of 30 degrees, 5 cm on an azimuth of 180 degrees, and 2 cm on an azimuth of 270 degrees. At that point, the observer loses patience and swats and ant. What is the X-Y position of the squashed ant, measured with respect of the place where it started?

SHOW WORK:

Initially it went 3cm, 30 degrees. This means the horizontal distance he travelled, indicated by the light blue line in the diagram, must be 3cos30 (since cos 30 = adjacent / hypotenuse). In the diagram we also see that the ant moves to the left 2cm, horizontally, at some point in time. To find the total horizontal displacement therefore, we subtract the rightward horizontal displacement by the leftward horizontal displacement:

3cos30 - 2 = 2.598 - 2 = 0.598cm

This means the ant is 0.598cm right of its original position. This is x in the diagram.

To figure out y, the vertical distance the ant has travelled, we do something similar. We need to find the difference between the downward displacement (5cm) and the upward displacement (3sin30 this time):

3sin30 - 5 = 1.5 - 5 = -3.5cm

The negative, in this case, denotes that the ant is below the original starting point. As a result, y = 3.5cm

Therefore, the final position is [0.598, 3.5].

ANSWER: 0.598, 3.5

3. A motorcycle rider is attempting to set a speed record on a straight course. The cycle passes two points, a measured 100 m apart, in 1.2 seconds. What is its average velocity between those points?

SHOW WORK: Velocity is the displacement over time. 100m / 1.2 = 83.3m/s [forward].

ANSWER: 83.3m/s [forward].

4. A M1911 pistol fires a .45 ACP round. The bullet leaves the 13 cm barrel with a velocity of 300 m/s.

Answer: the bullet is not moving at the beginning of the 13cm barrel, then leaves the barrel at 300m/s. The average velocity would be the average of all velocities while it was inside the length of the barrel, and we can take two points for this measurement: beginning of the barrel where velocity was zero, and end of barrel which was 300m/s.

Avg. velocity = (300m - 0) / 0.13m = 2310m/s

a. The time the bullet takes to travel through the 13cm barrel, exiting at 300m/s while starting at 0m/s, requires calculation using a classic kinematic equation: d = (v1 + v2) /2 * t

b. where v1 = initial velocity = 0 d = displacement = 13cm t = time 0.13m = (0 + 300m/s) / 2 * t t = 0.13 / 150 = 8.67x10^-4 seconds

c. Avg. acceleration: v2^2 = v1^2(t) + 2ad

300^2 = 0 + 2a(0.13) a = 300^2 / 0.26 = 3.46x10^5m/s

ANSWER1: 2310m/s

ANSWER2: -4 seconds

ANSWER3: 3.46x10^5m/s

5. A small powder charge propels a shoulder-launched antitank missile from its tube at 20 m/s. At 0.2 seconds after launch, the rocket motor ignites and accelerates the missile at a constant 20 g. After 2 seconds, the missile hits its target and detonates. How far away was the target? Assume a straight flight path. (Hint: Calculate two distances, one before and one after rocket motor ignition.)

Answer: distance traveled prior to rocket launch:

initial velocity = 20m/s

t = 0.2s Velocity = displacement / time

d = vt = 20*0.2 = 4m

distance after rocket ignition:

t = 2s

a = 20g

initial velocity = 20m/s

d = ?

d = v1(t) + 1/2 (a)(t^2)

d = 20(2) + 1/2 (20x9.81)(2^2)

d = 40 + 392.4 = 432.4m

*add the two distances together:

4 + 432.4 = 436.4m

ANSWER: 436.4m